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3 years ago

What needs to be corrected in this construction of a line parallel to AB passing through C?

Mathematics

2 answers:

Ierofanga [76]3 years ago

6 0

Answer: c

Step-by-step explanation:

lidiya [134]3 years ago

4 0

Answer:

I think it is c

Step-by-step explanation:

You might be interested in

A sinusoidal function whose period is 1/2 , maximum value is 10, and minimum value is −4 has a y-intercept of 3.

ValentinkaMS [17]

Ans: f(x)=7sin(4pix) + 3

We see the period, which is equivalent to 2pi divided by the coefficient of the argument of the trigonometric function, is 1/2 since 2pi/4i = 1/2

We see the maximum value of f(x) is 10 since sin(x) is bounded such that -1 < sin(x) < 1, therefore -7 < 7sin(x) < 7. And since we are adding 3 at the end of the equation, we can say the graph of 7sin(x) is shifted vertically 3 units, thus we have a max value of 10 and min value of -4 ( -4 < 7sin(x) + 3 < 10)

The y-intercept is seen as 3 since the sine function, at 0 radians i.e. x=0, has a value of 0 at the origin, this from the +3, we see the y-value of f(x) at the origin is 3.

8 0

3 years ago

Estimate.Solve problem<br><br> 3/6+1/3=

ycow [4]

Answer:

5/6

Step-by-step explanation:

5/6 would be the answer simplified I hope that helps :)

7 0

2 years ago

Read 2 more answers

How would I solve this?

Svetradugi [14.3K]

First ln on both sides to bring down the powers.
Ln and e cancel each other out.
3x +2 = 3
3x = 1
x = 1/3

6 0

3 years ago

The sum of the first 10 terms of an arithmetic series is 100 and the sum of next 10 300 .Find the series.

Free_Kalibri [48]

Let a be the first term in the sequence, and d the common difference between consecutive terms. If aₙ denotes the n-th term in the sequence, then

a₁ = a

a₂ = a₁ + d = a + d

a₃ = a₂ + d = a + 2d

a₄ = a₃ + d = a + 3d

and so on, up to the n-th term

aₙ = a + (n - 1) d

The sum of the first 10 terms is 100, and so

$\displaystyle \sum_{n=1}^{10} a_n = 100 \\ \sum_{n=1}^{10} (a + (n-1)d) = 100 \\ (a-d) \sum_{n=1}^{10} 1 + d \sum_{n=1}^{10} n = 100 \\ 10a+45d = 100$

where we use the well-known sum formulas,

$\displaystyle \sum_{n=1}^N 1 = 1 + 1 + 1 + \cdots + 1 = N$

$\displaystyle \sum_{n=1}^N n = 1 + 2 + 3 + \cdots + N = \frac{N(N+1)}2$

The sum of the next 10 terms is 300, so

$\displaystyle \sum_{n=11}^{20} a_n = 300 \\ (a-d) \sum_{n=11}^{20} 1 + d \sum_{n=11}^{20} n = 300 \\ (a-d) \left(\sum_{n=1}^{20} 1 - \sum_{n=1}^{10} 1\right) 1 + d \left(\sum_{n=1}^{20} n - \sum_{n=1}^{10} n\right) = 300 \\ 10a+145d = 300$

Solve for a and d. Eliminating a gives

(10a + 145d) - (10a + 45d) = 300 - 100

100d = 200

d = 2

and solving for a gives

10a + 145×2 = 300

10a = 10

a = 1

So, the given sequence is simply the sequence of positive odd integers,

{1, 3, 5, 7, 9, …}

given recursively by the relation

$\begin{cases}a_1 = 1 \\ a_n = a_{n-1} + 2 & \text{for }n>1\end{cases}$

and explicitly by

$a_n = 1 + 2(n-1) = 2n - 1$

for n ≥ 1.

7 0

2 years ago

Solve for X, (with steps please) <br> 0.46=0.527-[0.0562/2 * log(0.1+x/1.5-x)]

Ivanshal [37]

Rearrange the equation:

$\frac{0.0562}{2} * log( \frac{0.1+x}{1.5 - x}) = 0.527 - 0.46 = 0.067$

Isolate the log on its own

$log (\frac{0.1 + x}{1.5 - x}) = \frac{0.067*2}{0.0562} = 2.42 (2dp)$

To get rid of the log you must do 10^2.42

$\frac{0.1 +x}{1.5-x} = 10^{2.42} = 242.29 (2dp)$

Then multiply both sides by (1.5 - x)

$0.1 +x = 242.29(1.5-x)$

$0.1 + x = 363.44 - 242.29x$

Solve normally

$243.29x = 363.34 x = 363.34/243.29 x= 1.49344 x= 1.49 (2dp)$

Hope that helps. Please send me a message if there's patchy bits. Also I'm sure you'll figure this out, but just in case, anywhere I've put down a 2dp shows that I've rounded the number to 2 decimal places. It becomes a pain to deal with otherwise.

3 0

3 years ago